1 Introduction

Let \(X\) be a normed vector space, \(U\) be a bounded closed convex subset of \(X\), and \(f{:}\,X\rightarrow \bar{R}\) be a proper lower semicontinuous function. We define the perturbed minimal time function \(T^f{:}\,X\rightarrow R\) by

$$\begin{aligned} T^f(x):=\inf _{s\in X}\{T(x,s)+f(s)\},\quad \text {for all }x\in X, \end{aligned}$$
(1.1)

where \(T(x,s):=\inf \{t\ge 0{:}\,s-x\in tU\}.\) It is easy to see that, if \(U\equiv B\), then \(T(x-s)=\Vert x-s\Vert \), where \(B\) is the unit ball in \(X\).

For \(x\in X\), the perturbed minimal time problem is to find an element \(z_0\in X\) such that

$$\begin{aligned} T(x,z_0)+f(z_0)=T^f(x). \end{aligned}$$

In particular, if \(f=I_S\), where \(I_S\) denote the indicator function \(I_S\) of a closed set \(S\) (the definition will be given below), then the perturbed minimal time function \(T^f\) reduces to the minimal time function \(T_S\) in [14], which is defined by the following differential inclusion

$$\begin{aligned} \dot{x}(t)\in U,\quad x(0)=x. \end{aligned}$$
(1.2)

In other words,

$$\begin{aligned} T_S(x)\equiv {\left\{ \begin{array}{ll}\inf \{T>0{:}\, \text {there exists a trajectory }x(\cdot ) \text { satisfying } (1.2)\\ \text { with }x(0)=x\text { and }x(T)\in S\},&{}\quad x\notin S;\\ 0,&{}\quad x\in S.\end{array}\right. } \end{aligned}$$

If \(f=J+I_S\) and \(U\equiv B\), then the perturbed minimal time function \(T^f\) and the perturbed minimal time problem reduce to the perturbed distance function \(d_S^J\) and the perturbed optimization problem min\(_J(x,S)\) defined in [23], respectively, that is,

$$\begin{aligned} T^f(x)=d_S^J(x):=\inf _{s\in S}\{\Vert s-x\Vert +J(s)\},\quad \text {for all }x\in X, \end{aligned}$$

and

$$\begin{aligned} \text {min}_J(x,S):=\{z_0\in S| \Vert x-z_0\Vert +J(z_0)=d_S^J(x)\}. \end{aligned}$$

Baranger [1] proved that if \(S\) is a nonempty closed subset of a uniformly convex Banach space \(X\), then the set of all \(x\in X\) for the perturbed optimization problem min\(_J (x, S)\) has a solution is a dense \(G_\delta \)-subset of \(X\), which extends a result in [22] on the best approximation problem. For other results on perturbed optimization problems, see for example [3, 8, 9, 15, 16, 18–21]. In particular, Cobzas [9] extended Baranger’s result to the setting of reflexive Kadec Banach space; while Ni [18] relaxed the reflexivity assumption made in Cobzas’ result. The existence results have been applied to optimal control problems governed by partial differential equations, see for example, [1–3, 8, 12].

Assuming that the origin is an interior point of \(U\), Colombo and Wolenski [10, 11] studied the proximal and Fréchet subdifferentials of the function \(T_{S}(x)\) in a Hilbert space. He and Ng [13] studied the Fréchet and proximal subdifferentials of \(T_{S}\) in a Banach space. When the origin is an interior point of \(U\), the function \(T_{S}\) is globally Lipschitz, so the Clarke subdifferential of \(T_{S}\) is also discussed in [13]. Jiang and He [14] show the Frechét and proximal subdifferentials of the minimal time function \(T_{S}\) without requiring the origin be an interior point of \(U\) in normed space. In particular, if \(U\) is the (closed) unit ball in \(X\), then \(T_{S}(x)\) reduces to the usual distance \(d_S(x)\), which is defined by

$$\begin{aligned} d_S(x):=\inf _{s\in S}\Vert s-x\Vert \quad \text {for all }x\in X. \end{aligned}$$

The subdifferentials of \(d_S\) were studied in [4–7], and the subdifferentials of perturbed distance functions \(d_S^J\) were studied in [17, 23].

In order to reduce the symmetry of the norm, we replace the distance function in [23] by \(T(\cdot ,\cdot )\), which does not have the symmetry, and explore the Fréchet subdifferentials and the Proximal subdifferentials of its perturbed functions \(T^f(\cdot )\), the perturbed functions \(T^f(\cdot )\) are encountered in constraint optimization, via applying various perturbation, penalization, and approximation techniques. Our main results extend the corresponding ones in [14] from the minimal time function to perturbed minimal time function, and extend the corresponding ones in [23] from the general perturbed distance functions to general perturbed minimal time functions.

2 Preliminaries

Let \(X\) be a normed vector space with norm denoted by \(\Vert \cdot \Vert \). Let \(X^*\) denote the topological dual of \(X\). We use \(B(x;r)\) to denote the open ball centered at \(x\) with radius \(r>0\) and \(\langle \cdot ,\cdot \rangle \) to denote the pairing between \(X^*\) and \(X\). Let \(g{:}\,X\rightarrow \mathbb R \) be a lower semicontinuous function and \(x\in X\). \(g\) is said to be center Lipschitz on \(S\) at \(x\) with Lipschitz constant \(L\), if

$$\begin{aligned} |g(y)-g(x)|\le L\Vert y-x\Vert ,~~~\forall y\in S. \end{aligned}$$

Let us recall the following well-known classes of subdifferentials for \(g\) at \(x\).

  • The proximal subdifferential of \(g\) at \(x\) is the set

    $$\begin{aligned} \partial ^Pg(x):=\left\{ \xi \in X^*{:}\,\liminf _{\Vert v\Vert \rightarrow 0}\frac{g(x+v)-g(x)-\langle \xi ,v\rangle }{\Vert v\Vert ^2}>-\infty \right\} . \end{aligned}$$

    In other words, \(\xi \in \partial ^Pg(x)\) if and only if there exist \(\sigma >0\) and \(\delta >0\) such that

    $$\begin{aligned} g(x+v)-g(x)\ge \langle \xi ,v\rangle -\sigma \Vert v\Vert ^2,\quad \text {for all }v\in B(0,\delta ). \end{aligned}$$

  • The Frechét subdifferential of \(g\) at \(x\) is the set

    $$\begin{aligned} \partial ^{F}g(x):=\left\{ \xi \in X^*{:}\,\liminf _{\Vert v\Vert \rightarrow 0}\frac{g(x+v)-g(x)-\langle \xi ,v\rangle }{\Vert v\Vert }\ge 0\right\} . \end{aligned}$$

    That is, \(\xi \in \partial ^Fg(x)\) if and only if for any \(\sigma >0\), there exists \(\delta >0\) such that

    $$\begin{aligned} g(x+v)-g(x)\ge \langle \xi ,v\rangle -\sigma \Vert v\Vert ,\quad \text {for all }v\in B(0,\delta ). \end{aligned}$$

Recall that \(f\) satisfies the center Lipschitz condition on \(X\) at \(x\), if there exists \(L>0\) such that

$$\begin{aligned} |f(y)-f(x)|\le L\Vert y-x\Vert , \quad \text { for each }y\in X. \end{aligned}$$

The support function of a set \(K\subset X\) is defined by

$$\begin{aligned} \mathfrak I _{K}(\xi ):=\sup _{x\in K}\langle \xi ,x\rangle . \end{aligned}$$

The indicator function \(I_S\) of \(S\) is defined by

$$\begin{aligned} I_S(x)\equiv {\left\{ \begin{array}{ll}0,&{}x\in S;\\ +\infty ,&{}x\notin S.\end{array}\right. } \end{aligned}$$

In view of [14, Proposition 2.2], we have the following result.

Proposition 2.1

\(T(x,s)=0\) if and only if \(x=s\).

We use \(S_0\) to denote the set of all points \(x\in X\) such that \(x\) is a solution of the perturbed optimization problem, i.e.,

$$\begin{aligned} S_0=\{x\in X| T^f(x)=f(x)\}. \end{aligned}$$

Remark 2.1

It is obviously that, if \(f=I_S\), then \(S_0\) equals \(S\) in [14]; if \(f=I_S+J\) and \(U\) is the unit ball in \(X\), then \(S_0\) equals \(S_0\) in [23].

3 Fréchet subdifferential of a minimal time function

Theorem 3.1

Let \(x\in S_0\). The following assertions hold.

  1. 1.

    \(\partial ^F T^f(x)\subset \partial ^F f(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}.\)

  2. 2.

    If \(f(\cdot )\) is center Lipschitz on \(X\) at \(x\) with Lipschitz constant \(0\le L<1/M\), where \(M:=\sup _{u\in U}\Vert u\Vert \), then we have

    $$\begin{aligned} \partial ^F T^f(x)=\partial ^F f(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}. \end{aligned}$$

Proof

(1) Let \(\xi \in \partial ^F T^f(x)\). Then for any \(\sigma >0\), there exists \(\delta >0\) such that

$$\begin{aligned} T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert \end{aligned}$$
(3.1)

for all \(y\in B(x;\delta )\).

We will prove

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ,\quad \text { for all}\quad y\in B(x;\delta ). \end{aligned}$$
(3.2)

Thus \(\xi \in \partial ^Ff(x).\)

By (3.1) and definition of \(S_0\), (3.2) is trivial if \(y\in B(x;\delta )\cap S_0\), we may assume that \(y\in B(x;\delta ){\setminus } S_0\), by the definition of \(T^f\), we have \(T^f(y)\le f(y)\), and as \(x\in S_0\), we have from (3.1) that

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert \end{aligned}$$

Hence, \(f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert \), for all \(y\in B(x;\delta )\).

Fix any \(v\in U\). Let \(t_\lambda :=T^f(x-\lambda v)\), where \(\lambda >0\). Since \(x-(x-\lambda v)\in \lambda U\), \(T(x-\lambda v,x)\le \lambda ,\) \(t_\lambda \le \lambda +f(x)<\infty \). It follows from (3.1) that for sufficiently small \(\lambda >0\),

$$\begin{aligned} \lambda +f(x)\ge t_\lambda \ge f(x)+\lambda \langle -\xi ,v\rangle -\lambda \sigma \Vert v\Vert , \end{aligned}$$

which implies that \(\langle -\xi ,v\rangle \le 1+\sigma \Vert v\Vert \). Since \(\sigma >0\) and \(v\in U\) are arbitrary, \(\mathfrak I _{U}(-\xi )\le 1\).

(2) It is sufficient to prove

$$\begin{aligned} \partial ^Ff(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}\subset \partial ^F T^f(x). \end{aligned}$$

Let \(\xi \in \partial ^Ff(x)\) be such that \(\mathfrak I _{U}(-\xi )\le 1\).

For any \(\sigma >0\), take \(\sigma _0\in \left( 0,\frac{(1-LM)\sigma }{(1+M\Vert \xi \Vert )}\right) \). By the definition of Fréchet normal cone, there exists \(\delta >0\) such that

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma _0\Vert y-x\Vert ,\text { for all }y\in B(x;\delta ). \end{aligned}$$
(3.3)

Then

$$\begin{aligned} T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma _0\Vert y-x\Vert ,\text { for all }y\in S_0\cap B(x;\delta ).\quad \quad \end{aligned}$$
(3.4)

Let \(\delta _0:=\frac{(1-LM)\delta }{3(1+M\Vert \xi \Vert )}< \delta \). Then

$$\begin{aligned} T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma _0\Vert y-x\Vert ,\text { for all }y\in S_0\cap B(x;\delta _0).\quad \quad \end{aligned}$$
(3.5)

Now we prove that (3.5) holds for all \(y\in B(x;\delta _0){\setminus } S_0\). Therefore, \(\xi \in \partial ^FT^f(x)\).

If not, then there is \(y_0\not \in S_0\) such that

$$\begin{aligned} \Vert y_0-x\Vert < \delta _0\quad \text { and }\quad T^f(y_0)<T^f(x)+\langle \xi ,y_0-x\rangle -\sigma \Vert y_0-x\Vert . \end{aligned}$$
(3.6)

The latter implies that

$$\begin{aligned} T^f(y_0)\le f(x)+\Vert \xi \Vert \Vert y_0-x\Vert . \end{aligned}$$
(3.7)

Let \(t:=T^f(y_0)\). By the definition of \(T^f\), for any \({{\mathrm{\varepsilon }}}\in \left( 0,\frac{(1-LM)\delta }{3M}\right) \), there are \(t_1\in (0,t+{{\mathrm{\varepsilon }}})\), and \(s\in X\) such that \(t_1=T(y_0,s)+f(s)<t+{{\mathrm{\varepsilon }}}\). By the definition of \(T\), for any \({{\mathrm{\varepsilon }}}'\in \left( 0,\frac{(1-LM)\delta }{3M}\right) \), there are \(t_2\in (t_1-f(s),t_1-f(s)+{{\mathrm{\varepsilon }}}')\), \(u\in U\), such that \(s-y_0=t_2u.\) Thus (3.7) and \(f\) is center Lipshitz on \(X\) at \(x\) yield that

$$\begin{aligned} \Vert s-x\Vert&\le \Vert y_0-x\Vert +(t_1-f(s)+{{\mathrm{\varepsilon }}}')\Vert u\Vert \le \Vert y_0-x\Vert +(t+{{\mathrm{\varepsilon }}}-f(s)+{{\mathrm{\varepsilon }}}')M\\&\le \Vert y_0-x\Vert +(f(x)+\Vert \xi \Vert \Vert y_0-x\Vert +{{\mathrm{\varepsilon }}}-f(s)+{{\mathrm{\varepsilon }}}')M\\&\le (1+M\Vert \xi \Vert )\Vert y_0-x\Vert +(f(x)-f(s))M+({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')M\\&\le (1+M\Vert \xi \Vert )\Vert y_0-x\Vert +LM\Vert s-x\Vert +({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')M \end{aligned}$$

Then, we have

$$\begin{aligned} \Vert s-x\Vert \le \frac{1+M\Vert \xi \Vert }{1-LM}\Vert y_0-x\Vert +\frac{M}{1-LM}({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')<\delta . \end{aligned}$$
(3.8)

This verifies that \(s\in B(x;\delta )\). Applying (3.3), (3.8) and \(\mathfrak I _{U}(-\xi )\le 1\), we have

$$\begin{aligned}&T^f(y_0)-T^f(x)-\langle \xi ,y_0-x\rangle =t-f(x)-\langle \xi ,y_0-s\rangle -\langle \xi ,s-x\rangle \\&\quad \ge t-f(x)-\langle \xi ,y_0-s\rangle +(f(x)-f(s)-\sigma _0\Vert s-x\Vert )\\&\quad =t-(t_1-f(s)+{{\mathrm{\varepsilon }}}')\langle -\xi ,u\rangle -f(s)-\sigma _0\Vert s-x\Vert \\&\quad \ge t-(t_1-f(s)+{{\mathrm{\varepsilon }}}')-f(s)-\sigma _0\Vert s-x\Vert \\&\quad \ge -{{\mathrm{\varepsilon }}}-{{\mathrm{\varepsilon }}}'-\sigma _0\Vert s-x\Vert \\&\quad \ge -{{\mathrm{\varepsilon }}}-{{\mathrm{\varepsilon }}}'-\sigma _0 \left( \frac{1+M\Vert \xi \Vert }{1-LM}\Vert y_0-x\Vert +\frac{M}{1-LM}({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')\right) \\&\quad \ge -\left( 1+\frac{M}{1-LM}\sigma _0\right) ({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')-\frac{1+M\Vert \xi \Vert }{1-LM}\sigma _0\Vert y_0-x\Vert \\&\quad \ge -\left( 1+\frac{M}{1-LM}\sigma _0\right) ({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')-\sigma \Vert y_0-x\Vert . \end{aligned}$$

Letting \({{\mathrm{\varepsilon }}}'\rightarrow 0\!+\text { and }{{\mathrm{\varepsilon }}}\rightarrow 0+\), it yields that

$$\begin{aligned} T^f(y_0)-T^f(x)-\langle \xi ,y_0-x\rangle \ge -\sigma \Vert y_0-x\Vert , \end{aligned}$$

which contradicts to (3.6). \(\square \)

In particular, letting \(f=I_S\), we get the following corollary, which was proved in [14].

Corollary 3.1

Assume that \(f=I_S\), where \(S\) is a closed convex subset of \(X\), if \(x\in S\), then

$$\begin{aligned} \partial ^F T^f(x)=\partial ^F T(x)=N_{S}^F(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}. \end{aligned}$$

In particular, letting \(f=I_S+J\) and \(U\equiv B\), we get the following corollary, which was proved in [23].

Corollary 3.2

Assume that \(f=I_S+J\) and \(U\equiv B\), where \(B\) is the unit ball in \(X\) and \(S\) is a closed convex subset of \(X\), let \(x\in S_0\). The following assertions hold.

  1. 1.

    \(\partial ^F T^f(x)=\partial ^F d_S^J(x)\subset \partial ^F(J+I_S)(x)\cap B^*.\)

  2. 2.

    If \(J(\cdot )\) is center Lipschitz on \(S\) at \(x\) with Lipschitz constant \(0\le L<1\), then we have

    $$\begin{aligned} \partial ^F T^f(x)=\partial ^F d_S^J(x)=\partial ^F(J+I_S)(x)\cap B^*. \end{aligned}$$

4 Proximal subdifferential of a minimal time function

Theorem 4.1

Let \(x\in S_0\). The following assertions hold.

  1. 1.

    \(\partial ^P T^f(x)\subset \partial ^Pf(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}.\)

  2. 2.

    If \(f(\cdot )\) is center Lipschitz on \(X\) at \(x\) with Lipschitz constant \(0\le L<1/M\), where \(M:=\sup _{u\in U}\Vert u\Vert \), then we have

    $$\begin{aligned} \partial ^P T^f(x)=\partial ^Pf(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}. \end{aligned}$$

Proof

(1) Let \(\xi \in \partial ^P T^f(x)\). Then there exist \(\sigma ,\delta >0\) such that

$$\begin{aligned} T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2 \end{aligned}$$
(4.1)

for all \(y\in B(x;\delta )\).

We wil prove

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2,\text { for all}\quad y\in B(x;\delta ). \end{aligned}$$
(4.2)

Then \(\xi \in \partial ^Pf(x).\)

By (4.1) and the definition of \(S_0\), (4.2) is trivial if \(y\in B(x;\delta )\cap S_0\), we may assume that \(y\in B(x;\delta )\setminus S_0\). By the definition of \(T^f\), we have \(T^f(y)\le f(y)\), and as \(x\in S_0\), we have from (4.1) that

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2. \end{aligned}$$

Hence, \(f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2\), for all \(y\in B(x;\delta )\).

Fix any \(v\in U\). Let \(t_\lambda :=T^F(x-\lambda v)\), where \(\lambda >0\). Since \(x-(x-\lambda v)\in \lambda U\), \(T(x-\lambda v,x)\le \lambda ,\) \(t_\lambda \le \lambda +f(x)<\infty \). It follows from (4.1) that for sufficiently small \(\lambda >0\),

$$\begin{aligned} \lambda +f(x)\ge t_\lambda \ge f(x)+\lambda \langle -\xi ,v\rangle -\lambda ^2\sigma \Vert v\Vert ^2, \end{aligned}$$

which implies that \(\langle -\xi ,v\rangle \le 1\). Therefore, \(\mathfrak I _{U}(-\xi )\le 1\).

(2) It is sufficient to prove

$$\begin{aligned} \partial ^Pf(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}\subset \partial ^P T^f(x). \end{aligned}$$

Let \(\xi \in \partial ^Pf(x)\) be such that \(\mathfrak I _{U}(-\xi )\le 1\). Then there exist \(\sigma _1,\delta >0\) such that

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma _1\Vert y-x\Vert ^2,\text { for all }y\in B(x;\delta ). \end{aligned}$$
(4.3)

Take \(\sigma :=2\left( \frac{1+M\Vert \xi \Vert }{1-LM}\right) ^2\sigma _1>\sigma _1\). Thus (4.3) implies that

$$\begin{aligned} f(y)-f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2,\text { for all }y\in B(x;\delta ). \end{aligned}$$
(4.4)

and

$$\begin{aligned} T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2,\text { for all }y\in S_0\cap B(x;\delta ).\quad \quad \end{aligned}$$
(4.5)

Let \(\delta _0:=\frac{(1-LM)\delta }{3(1+M\Vert \xi \Vert )}< \delta \). Then

$$\begin{aligned} T^f(y)-T^f(x)-\langle \xi ,y-x\rangle \ge -\sigma \Vert y-x\Vert ^2,\text { for all }y\in S_0\cap B(x;\delta _0).\quad \quad \end{aligned}$$
(4.6)

Now we prove that (4.6) holds for all \(y\in B(x;\delta _0){\setminus } S_0\). Therefore, \(\xi \in \partial ^PT^f(x)\).

If not, then there is \(y_0\not \in S_0\) such that

$$\begin{aligned} \Vert y_0-x\Vert < \delta _0\quad \text { and }\quad T^f(y_0)<T^f(x)+\langle \xi ,y_0-x\rangle -\sigma \Vert y_0-x\Vert ^2. \end{aligned}$$
(4.7)

The latter implies that

$$\begin{aligned} T^f(y_0)\le J(x)+\Vert \xi \Vert \Vert y_0-x\Vert . \end{aligned}$$
(4.8)

Let \(t:=T^f(y_0)\). By the definition of \(T^f\), for any \({{\mathrm{\varepsilon }}}\in \left( 0,\frac{(1-LM)\delta }{3M}\right) \), there are \(t_1\in (t,t+{{\mathrm{\varepsilon }}})\), and \(s\in X\) such that \(t_1=T(y_0,s)+f(s)<t+{{\mathrm{\varepsilon }}}\), by the definition of \(T\), for any \({{\mathrm{\varepsilon }}}'\in \left( 0,\frac{(1-LM)\delta }{3M}\right) \), there are \(t_2\in (t_1-f(s),t_1-f(s)+{{\mathrm{\varepsilon }}}')\) \(u\in U\), such that \(s-y_0=t_2u.\) Thus (4.8) and \(f\) is center Lipshitz on \(X\) at \(x\) yield that

$$\begin{aligned} \Vert s-x\Vert&\le \Vert y_0-x\Vert +(t_1-f(s)+{{\mathrm{\varepsilon }}}')\Vert u\Vert \le \Vert y_0-x\Vert +(t+{{\mathrm{\varepsilon }}}-f(s)+{{\mathrm{\varepsilon }}}')M\\&\le \Vert y_0-x\Vert +(f(x)+\Vert \xi \Vert \Vert y_0-x\Vert +{{\mathrm{\varepsilon }}}-f(s)+{{\mathrm{\varepsilon }}}')M\\&\le (1+M\Vert \xi \Vert )\Vert y_0-x\Vert +(f(x)-f(s))M+({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')M\\&\le (1+M\Vert \xi \Vert )\Vert y_0-x\Vert +LM\Vert s-x\Vert +({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')M\\ \end{aligned}$$

Then, we have

$$\begin{aligned} \Vert s-x\Vert \le \frac{1+M\Vert \xi \Vert }{1-LM}\Vert y_0-x\Vert +\frac{M}{1-LM}({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')<\delta . \end{aligned}$$
(4.9)

This verifies that \(s\in B(x;\delta )\). Applying (4.4), (4.9) and \(\mathfrak I _{U}(-\xi )\le 1\), we have

$$\begin{aligned}&T^f(y_0)-T^f(x)-\langle \xi ,y_0-x\rangle =t-f(x)-\langle \xi ,y_0-s\rangle -\langle \xi ,s-x\rangle \\&\quad \ge t-f(x)-\langle \xi ,y_0-s\rangle +(f(x)-f(s)-\sigma _1\Vert s-x\Vert ^2)\\&\quad =t-(t_1-f(s)+{{\mathrm{\varepsilon }}}')\langle -\xi ,u\rangle -f(s)-\sigma _1\Vert s-x\Vert ^2\\&\quad \ge t-(t_1-f(s)+{{\mathrm{\varepsilon }}}')-f(s)-\sigma _1\Vert s-x\Vert ^2\\&\quad \ge -{{\mathrm{\varepsilon }}}-{{\mathrm{\varepsilon }}}'-\sigma _1\Vert s-x\Vert ^2\\&\quad \ge -{{\mathrm{\varepsilon }}}-{{\mathrm{\varepsilon }}}'-2\sigma _1 \left( \frac{1+M\Vert \xi \Vert }{1-LM}\right) ^2\Vert y_0-x\Vert ^2 -2\sigma _0\left( \frac{M}{1-LM}\right) ^2({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')^2\\&\quad \ge -{{\mathrm{\varepsilon }}}-{{\mathrm{\varepsilon }}}'-2\sigma _1\left( \frac{M}{1-LM}\right) ^2({{\mathrm{\varepsilon }}}+{{\mathrm{\varepsilon }}}')^2-\sigma \Vert y_0-x\Vert ^2. \end{aligned}$$

Letting \({{\mathrm{\varepsilon }}}'\rightarrow 0+\text { and }{{\mathrm{\varepsilon }}}\rightarrow 0+\), it yields that

$$\begin{aligned} T^f(y_0)-T^f(x)-\langle \xi ,y_0-x\rangle \ge -\sigma \Vert y_0-x\Vert ^2, \end{aligned}$$

which contradicts to (4.7). \(\square \)

In particular, letting \(f=I_S\), we get the following corollary, which is proved in [14].

Corollary 4.1

Assume that \(f=I_S\), where \(S\) is a closed convex subset of \(X\), if \(x\in S\), then

$$\begin{aligned} \partial ^P T^f(x)=\partial ^P T(x)=N_{S}^P(x)\cap \{\xi \in X^*{:}\,\mathfrak I _{U}(-\xi )\le 1\}. \end{aligned}$$

In particular, letting \(f=I_S+J\) and \(U\equiv B\), we get the following corollary, which was proved in [23].

Corollary 4.2

Assume that \(f=I_S+J\) and \(U\equiv B\), where \(B\) is the unit ball in \(X\) and \(S\) is a closed convex subset of \(X\), let \(x\in S_0\). The following assertions hold.

  1. 1.

    \(\partial ^P T^f(x)=\partial ^P d_S^J(x)\subset \partial ^P(J+I_S)(x)\cap B^*.\)

  2. 2.

    If \(J(\cdot )\) is center Lipschitz on \(S\) at \(x\) with Lipschitz constant \(0\le L<1\), then we have

    $$\begin{aligned} \partial ^P T^f(x)=\partial ^P d_S^J(x)=\partial ^P(J+I_S)(x)\cap B^*. \end{aligned}$$